complex stuff (1 Viewer)

[cutemouse]

How do you use that method so seamlessly? Are there tricks or do you just have to practise?

It's from the Terry Lee book. Don't use it, it's way too confusing and lots of room for error.

Plus it's not like you get a whole list of questions on this. In a test you'll probs only get like 2 marks worth, so it's not worth stuffing up by using some stupid shortcut.

 

[bob fossil]

It's from the Terry Lee book. Don't use it, it's way too confusing and lots of room for error.

Plus it's not like you get a whole list of questions on this. In a test you'll probs only get like 2 marks worth, so it's not worth stuffing up by using some stupid shortcut.

Doesn't the normal way take about 45 seconds?

 

[Aldehyde]

im from qld and we had to use de moivres theorem to simplify radicals:

(a+bi)^n = (rcis(theta))^n = r^ncis(n(theta)) = r^n[cos(n*theta) + isin(n*theta)]


z = (-12 - 16i) ^1/2

r = sqrt((-12)^2 + (-16)^2) = sqrt (400) = 20

theta = tan^-1 (-16/-12) = tan^-1(4/3) [3rd quadrant, so add pi]
-> theta = pi + tan^-1(4/3)

then: -12 - 16i = (20cis(pi + tan^-1(4/3) ))^1/2

to get all solutions: (a+bi)^1/n = (rcis(theta))^1/n = r^1/ncis((theta + 2kpi/n) = r^1/n[cos(theta +2kpi/n) + isin(theta + 2kpi/n)] , for k from 0 to n-1

 

[cutemouse]

Yes, you can use DeMovire's theorem to find square roots of equations, since DeMovire's theorem is true for all real values of n. Too much fiddling around in my opinion though.

 

[The Nomad]

Yes, you can use DeMovire's theorem to find square roots of equations, since DeMovire's theorem is true for all real values of n. Too much fiddling around in my opinion though.

And it isn't always the greatest when the roots are required in the form x + iy.

 

[bob fossil]

I failed the 3 unit exam with only getting 34% so the pressure was on to perform in 4 unit.


Fuck yeah I passed with 56%.



Thanks to everyone that answered my questions.